The locus of the point of intersection of the lines $$\sqrt{2}x - y + 4\sqrt{2}k = 0$$ and $$\sqrt{2}kx + ky - 4\sqrt{2} = 0$$ (k is any non-zero real parameter) is:

We have two concurrent straight lines depending on a parameter $$k\neq 0$$:

$$\sqrt{2}\,x-y+4\sqrt{2}\,k=0\qquad\text{and}\qquad \sqrt{2}\,k\,x+k\,y-4\sqrt{2}=0.$$

The task is to find the locus of their point of intersection. Let the required point be $$(x,y)$$. Because the point lies on the first line, we may express $$y$$ in terms of $$x$$ and $$k$$:

$$\sqrt{2}\,x-y+4\sqrt{2}\,k=0\;\Longrightarrow\;y=\sqrt{2}\,x+4\sqrt{2}\,k.$$

Now we substitute this value of $$y$$ in the second line:

$$\sqrt{2}\,k\,x+k\bigl(\sqrt{2}\,x+4\sqrt{2}\,k\bigr)-4\sqrt{2}=0.$$

Simplifying term by term,

$$\sqrt{2}\,k\,x+k\sqrt{2}\,x+4\sqrt{2}\,k^{2}-4\sqrt{2}=0,$$

and the first two terms are identical, so they combine to give

$$2\sqrt{2}\,k\,x+4\sqrt{2}\,k^{2}-4\sqrt{2}=0.$$

Dividing the whole equation by $$2\sqrt{2}$$ (non-zero) yields

$$k\,x+2k^{2}-2=0.$$

This is a quadratic in $$k$$:

$$2k^{2}+xk-2=0.$$

For a quadratic $$ak^{2}+bk+c=0$$, the quadratic formula gives $$k=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Applying it with $$a=2,\; b=x,\; c=-2$$ we obtain

$$k=\dfrac{-x\pm\sqrt{x^{2}+16}}{4}.$$

Next we substitute this value of $$k$$ back in the earlier expression for $$y$$:

$$y=\sqrt{2}\,x+4\sqrt{2}\,k =\sqrt{2}\,x+4\sqrt{2}\left(\dfrac{-x\pm\sqrt{x^{2}+16}}{4}\right) =\sqrt{2}\,x+\sqrt{2}\bigl(-x\pm\sqrt{x^{2}+16}\bigr).$$

The $$\sqrt{2}\,x$$ terms cancel, leaving

$$y=\pm\sqrt{2}\,\sqrt{x^{2}+16}.$$

Eliminating the ambiguous sign by squaring, we get

$$y^{2}=2\bigl(x^{2}+16\bigr)=2x^{2}+32.$$

Rearranging the terms,

$$y^{2}-2x^{2}=32.$$

To recognise the conic, we divide by $$32$$:

$$\frac{y^{2}}{32}-\frac{x^{2}}{16}=1.$$

This is clearly of the standard form $$\dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1,$$ which represents a hyperbola whose transverse axis is along the $$y$$-axis. Here

$$a^{2}=32\;\Longrightarrow\;a=\sqrt{32}=4\sqrt{2}.$$

The length of the transverse axis of a hyperbola is $$2a$$, so in the present case

$$2a=2\bigl(4\sqrt{2}\bigr)=8\sqrt{2}.$$

Among the given options, this matches the statement “a hyperbola with length of its transverse axis $$8\sqrt{2}$$.”

Hence, the correct answer is Option C.